D and E are the ones that aren’t functions because they use the x-values more then once
Proving a relation for all natural numbers involves proving it for n = 1 and showing that it holds for n + 1 if it is assumed that it is true for any n.
The relation 2+4+6+...+2n = n^2+n has to be proved.
If n = 1, the right hand side is equal to 2*1 = 2 and the left hand side is equal to 1^1 + 1 = 1 + 1 = 2
Assume that the relation holds for any value of n.
2 + 4 + 6 + ... + 2n + 2(n+1) = n^2 + n + 2(n + 1)
= n^2 + n + 2n + 2
= n^2 + 2n + 1 + n + 1
= (n + 1)^2 + (n + 1)
This shows that the given relation is true for n = 1 and if it is assumed to be true for n it is also true for n + 1.
<span>By mathematical induction the relation is true for any value of n.</span>
Answer:
B
Step-by-step explanation:
Equation of line 1:
Choose two points : (-1, 0) & (0,2)
y -intercept = b = 2
y = mx+ 2
Plugin the values of the points ( -1 , 0) in the above equation
0 = -1m + 2
-2 = -m
m = 2
Equation of line 1 : y = 2x + 2
Equation of line 2:
(5,0) & (0,5)
y-intercept = b = 5
y = mx +b
y = mx + 5
Plugin the value of points (5 , 0) in the above equation
0 = 5m + 5
-5 = 5m
-5/5 = m
m = -1
Equation of line 2: y = -x + 5
Conclusion: 2x + 2 = -x + 5
